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mamaluj [8]
3 years ago
7

an object which has a mass of 70 kg is sitting on a cliff 10 m high calculate the objects gravitational potential energy given g

= 1o/s^
Physics
1 answer:
Alex73 [517]3 years ago
5 0
GPE = mgh (for objects close to earth where g can be considered to be constant) Where m is the mass, g is the acceleration due to gravity (which the question has told us to assume is 10ms^-2), and h is the height of the object.

Substituting the values in:
GPE = 70*10*10 = 7000J
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Show how the alternative definition of power, found in your book, can be derived by substituting the definitions of work and spe
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Let us consider body moves a distance S due to the force F.

Hence the work by the body W = FS

If the force is not along the direction of displacement,then the work by a body for travelling a distance S will be -

                                       W=[ Fcos\theta]*S  where    Fcos\theta is the component of the force along the direction of displacement.

                                  Hence\ W= FScos\theta

                                                        = F.S

As per the question the power P is given as -

                                                  P=\frac{W}{\delta t}

                                                         =\frac{F.S}{\delta t}

                                                         = F.\frac{S}{\delta t}

                                                         = \ F.V

Hence alternative definition of power P = F.V


8 0
3 years ago
Read 2 more answers
What happens during the process of deposition
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Deposition is the process in which sediments, soil and rocks are added to a landform or landmass. When previous weathers surface material , is deposited to a building layer of sediment .
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The soccer game begins with a ______ ______ in the center of the field
blagie [28]
It begins with a kick-off
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3 years ago
10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when
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Answer:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

Explanation:

For this case  we can use the second law of Newton given by:

\sum F = ma

The friction force on this case is defined as :

F_f = \mu_k N = \mu_k mg

Where N represent the normal force, \mu_k the kinetic friction coeffient and a the acceleration.

For this case we can assume that the only force is the friction force and we have:

F_f = ma

Replacing the friction force we got:

\mu_k mg = ma

We can cancel the mass and we have:

a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}

And now we can use the following kinematic formula in order to find the distance travelled:

v^2_f = v^2_i - 2ad

Assuming the final velocity is 0 we can find the distance like this:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

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3 years ago
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Answer:

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3 years ago
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